Search: a111035 -id:a111035
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1, 6479, 11663, 34943, 47519, 51983, 196559, 327359, 685583, 954239, 1016063, 1346879, 2039183, 2332799, 2504447, 4665599, 5143823, 5962319, 6128639, 6723359, 7225343, 9363599, 12027023, 12446783, 14930351, 17639999, 17735759, 22924943, 24681023, 34715519, 41990399
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OFFSET
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1,2
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LINKS
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PROG
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(PARI) f(n, m) = (Mod([1, 1; 1, 0], m)^n)[1, 2];
isok(n) = f(n+2, n)==1 && f(n+3, n+1)==1;
for(k=1, 10^7, if(isok(k), print1(k, ", "))); \\ Daniel Suteu, Feb 03 2020
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CROSSREFS
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Cf. A111035 (the sum of the first k nonzero Fibonacci numbers is divisible by k).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A101907
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Numbers n-1 such that the arithmetic mean of the first n Fibonacci numbers (beginning with F(0)) is an integer.
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+10
6
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0, 3, 5, 8, 10, 18, 23, 28, 30, 33, 40, 45, 47, 58, 60, 70, 71, 78, 88, 93, 95, 99, 100, 105, 108, 119, 128, 130, 138, 143, 148, 150, 165, 178, 180, 190, 191, 198, 200, 210, 213, 215, 219, 225, 228, 238, 239, 240, 248, 250, 268, 270, 273, 280, 287, 310, 320, 330
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OFFSET
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1,2
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COMMENTS
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The sum of the first n Fibonacci numbers is F(n+2)-1, sequence A000071.
Knott discusses the factorization of these numbers. - T. D. Noe, Oct 10 2005
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LINKS
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Eric W. Weisstein's World of Mathematics, Fibonacci
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FORMULA
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Numbers n-1 such that (F(0)+ F(1)+ ... + F(n-1)) / n is an integer. F(i) is the i-th Fibonacci number.
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EXAMPLE
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n=4 : (F(0)+F(1)+F(2)+F(3))/4 = (0+1+1+2)/4 = 1. So n-1 = 4-1 = 3 is a term.
n=6 : (F(0)+F(1)+F(2)+F(3)+F(4)+F(5))/6 = (0+1+1+2+3+5)/6 = 2. So n-1 = 6-1 = 5 is a term.
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A124456
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Numbers n which divide the sum of the Fibonacci numbers F(1) through F(n) and such that n is not a multiple of 24.
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+10
6
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1, 2, 77, 319, 323, 1517, 3021, 4757, 6479, 7221, 8159, 8229, 9797, 11663, 12597, 13629, 13869, 14429, 14949, 16637, 18407, 19043, 19437, 23407, 24947, 25437, 30049, 30621, 34943, 34989, 35207, 39203, 43677, 44099, 47519, 51983, 53663, 55221, 65471, 70221, 77837, 78089, 79547
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OFFSET
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1,2
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COMMENTS
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Numbers n which divide the sum of the first n nonzero Fibonacci numbers are listed in A111035 = {1, 2, 24, 48, 72, 77, 96, ...}. Most of these are multiples of 24. These multiples divided by 24 are listed in A124455 = {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ...}. [Edited by M. F. Hasler, Feb 04 2020]
The even terms a({2, 155, 397, 469, ...}) = {2, 758642, 7057466, 10805846, ...} are now listed in A331870. - M. F. Hasler, Feb 06 2020
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LINKS
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FORMULA
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MATHEMATICA
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Select[Range[20000], !IntegerQ[ #/24]&&Mod[Fibonacci[ #+2]-1, # ]==0&]
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PROG
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(Sage) [n for n in (1..20000) if mod(n, 24)!=0 and mod(fibonacci(n+2)-1, n)==0 ] # G. C. Greubel, Feb 16 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A331976
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Odd integers m that divide the sum of the first m nonzero Fibonacci numbers.
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+10
4
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1, 77, 319, 323, 1517, 3021, 4757, 6479, 7221, 8159, 8229, 9797, 11663, 12597, 13629, 13869, 14429, 14949, 16637, 18407, 19043, 19437, 23407, 24947, 25437, 30049, 30621, 34943, 34989, 35207, 39203, 43677, 44099, 47519, 51983, 53663, 55221, 65471, 70221, 77837, 78089, 79547
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OFFSET
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1,2
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COMMENTS
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Not all terms are squarefree, for instance 13869, 14949, 43677, 93357, ... are not.
A subsequence of A124456, missing just the even terms A124456({2, 155, 397, 469, ...}) = {2, 758624, 7057466, 10805846, ...}. - M. F. Hasler, Feb 06 2020
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LINKS
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FORMULA
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A111058
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Numbers k such that the average of the first k Lucas numbers is an integer.
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+10
3
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1, 2, 8, 12, 20, 24, 48, 60, 68, 72, 92, 96, 120, 140, 144, 188, 192, 200, 212, 216, 240, 288, 300, 332, 336, 360, 384, 428, 432, 440, 452, 480, 500, 548, 576, 600, 648, 660, 668, 672, 680, 692, 696, 720, 768, 780, 788, 812, 864, 908, 932, 960, 1008, 1028, 1052
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OFFSET
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1,2
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COMMENTS
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A111035 is the equivalent for Fibonacci numbers and has many elements in common with this sequence. T. D. Noe, who extended this sequence, noticed that, for some reason, 24 divides many of those k.
All terms are even except for the first term. - Harvey P. Dale, Apr 22 2024
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LINKS
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FORMULA
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k such that (Sum_{i=1..k} A000204(i))/k is an integer.
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MATHEMATICA
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Lucas[n_] := Fibonacci[n+1]+Fibonacci[n-1]; lst={}; s=0; Do[s=s+Lucas[n]; If[Mod[s, n]==0, AppendTo[lst, n]], {n, 1000}]; lst (* T. D. Noe *)
Module[{nn=1000, ln}, ln=LucasL[Range[nn]]; Table[If[IntegerQ[Mean[Take[ln, n]]], n, Nothing], {n, nn}]] (* Harvey P. Dale, Apr 22 2024 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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A331870
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Even numbers n which divide the sum of the Fibonacci numbers F(1) + ... + F(n) but are not a multiple of 24.
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+10
3
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2, 758642, 7057466, 10805846, 50860946, 59677526, 61800878, 155045678, 178551374, 217281146, 343943882, 359455694, 432175586, 609069506, 1449599486, 1721358698, 1829675354, 1884592706, 2013264194, 2116706282, 2680549946, 2971193186, 3084402122, 3252387386, 3454785386
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OFFSET
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1,1
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COMMENTS
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A111035 lists numbers n which divide the sum of the first n nonzero Fibonacci numbers. Most of these are multiples of 24. Sequence A124456 lists those which aren't. Most of these are odd (cf. A331976), this sequence lists the exceptions.
If we consider F(n+2) = 1 + the sum of the first n nonzero Fibonacci numbers (cf. A000071), then for even n we find:
4 divides F(n+2) for n == 4 (mod 12), 3 divides F(n+2) for n == 6 (mod 12),
F(n+2) == 3 (mod 4) for n == 8 (mod 12), 2 divides F(n+2) for n == 10 (mod 12),
F(n+2) == 5 (mod 6) for n == 12 (mod 24).
These relations imply that all terms a(n) == 2 (mod 12) for all n. This also means that all terms of A111035 are either divisible by 24, or odd, or congruent to 2 (mod 12).
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LINKS
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FORMULA
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a(n) == 2 (mod 12) for all n.
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PROG
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(PARI) M=[1, 1; 1, 0]; forstep(n=2, oo, 12, n%24&&(Mod(M, n)^(n+1))[1, 1]==1&& print1(n", ")) \\ Custom implementation of is_A111035(), check for updates there.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A124455
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Numbers n such that 24n divides the sum of the first 24n nonzero Fibonacci numbers.
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+10
2
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1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 24, 25, 27, 28, 30, 32, 36, 40, 42, 45, 46, 48, 50, 51, 54, 55, 56, 57, 60, 64, 70, 72, 75, 80, 81, 84, 86, 90, 92, 96, 98, 100, 102, 108, 110, 112, 114, 120, 125, 126, 128, 135, 138, 140, 144, 150, 153, 155, 160, 162
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OFFSET
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1,2
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COMMENTS
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Numbers n such that n divides the sum of the first n nonzero Fibonacci numbers are listed in A111035 = {1, 2, 24, 48, 72, 77, 96, 120, 144, 192, 216, 240, 288, 319, 323, 336, 360, ...}. Most of these are multiples of 24. Those which are not a multiple of 24 are listed in A124456 = {1, 2, 77, 319, 323, 1517, 3021, 4757, 6479, 7221, 8159, 8229, 9797, ...}.
This sequence coincides with A072378 (12n | F(12n)) for all values up to 84. The first two different terms are 86 and 164.
Prime a(n) are {2, 3, 5, 281, ...}.
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LINKS
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MATHEMATICA
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Select[Range[10000], IntegerQ[ #/24]&&Mod[Fibonacci[ #+2]-1, # ]==0&] /24
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A160757
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Averages of the Fibonacci numbers which take integer values.
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+10
1
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1, 1, 5058, 262213938, 18124577012898, 187952389930860, 1409394295257361938, 116903055445824294157698, 10100618828005365858877129458, 81435914480042681825934186407384633298, 7505278652741640947693896415563573183203138, 700346071081054203480884565881868806176873272498
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OFFSET
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1,3
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COMMENTS
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The numbers n such that F(1)+F(2)+...+F(n)=F(n+2)-1 is divisible by n are given in A111035. [From Max Alekseyev, Jun 04 2009]
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LINKS
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FORMULA
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1/n*Sum {j=1..n} Fibonacci_j is an integer.
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MATHEMATICA
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lst = {}; Do[a = Sum[ Fibonacci@ j, {j, n}]/n; If[ IntegerQ@ a, AppendTo[lst, a]], {n, 250}]; lst
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A282772
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Starting from F(n), minimum number, greater than 1, of consecutive Fibonacci numbers whose average is an integer.
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+10
1
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4, 2, 3, 12, 2, 13, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 12, 4, 2, 3, 12, 2, 5, 3, 2, 6, 13, 2, 12, 4, 2, 3, 5, 2, 24, 3, 2, 5, 24, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 5, 4, 2, 3, 12, 2, 24, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6
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OFFSET
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0,1
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COMMENTS
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Entries are 2, 3, 4, 5, 6, 12, 13 and 24.
Periodic with period equal to 420.
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LINKS
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FORMULA
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a(3*k + 1) = 2;
a(12*k + 2) = a(12*k + 6) = 3;
a(12*k) = 4;
a(30*k + 9) = a(30*k + 29) = a(60*k + 44) = 5;
a(60*k + 8) = a(60*k + 20) = a(60*k + 32) = a(60*k + 56) = 6;
a(60*k + 3) = a(60*k + 11) = a(60*k + 15) = a(60*k + 23) = a(60*k + 27) = a(60*k + 35) = a(60*k + 47) = a(60*k + 51) = 12;
a(420*k + 5) = a(420*k + 33) = a(420*k + 117) = a(420*k + 173) = a(420*k + 201) = a(420*k + 257) = a(420*k + 285) = a(420*k + 341) = 13;
a(420*k + x) = 24, with x = 17, 21, 41, 45, 53, 57, 65, 77, 81, 93, 101, 105, 113, 125, 137, 141, 153, 161, 165, 177, 185, 197, 213, 221, 225, 233, 237, 245, 261, 273, 281, 293, 297, 305, 317, 321, 333, 345, 353, 365, 377, 381, 393, 401, 405, 413, 417.
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EXAMPLE
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a(0) = 4 because F(0) + F(1) + F(2) + F(3) = 0 + 1 + 1 + 2 = 4 and 4/4 = 1;
a(1) = 2 because F(1) + F(2) = 1 + 1 = 2 and 2/2 = 1;
a(2) = 3 because F(2) + F(3) + F(4) = 1 + 2 + 3 = 6 and 6/3 = 2;
a(3) = 12 because F(3) + F(4) + ... + F(13) + F(14) = 2 + 3 + ... + 233 + 377 = 984 and 984/12 = 82.
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MAPLE
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with(combinat): P:=proc(q) local a, k, n; for k from 0 to q do a:=fibonacci(k); for n from 1 to q do a:=a+fibonacci(k+n);
if type(a/(n+1), integer) then print(n+1); break; fi; od; od; end: P(10^3);
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MATHEMATICA
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Table[k = 1; While[! IntegerQ@ Mean@ Take[#, n ;; n + k], k++]; k + 1, {n, Length@ # - 24}] &@ Fibonacci@ Range[0, 419] (* Michael De Vlieger, Mar 06 2017 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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OFFSET
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1,2
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COMMENTS
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a(1)-a(5) are given on p. 11 in Yaqubi, Fatehizadeh, 2020. According to the authors there are no other terms up to 10^6.
Apparently an erroneous version of A331977.
Included in accordance with OEIS policy of including published but erroneous sequences to serve as pointers to the correct values.
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LINKS
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PROG
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(PARI) is(n) = my(s=sum(i=1, n, fibonacci(i))); lift(Mod(s, n))==0 && lift(Mod(s+fibonacci(n+1), n+1))==0
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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STATUS
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approved
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